Does constexpr imply inline?

Question

Consider the following inlined function :

// Inline specifier version
#include<iostream>
#include<cstdlib>

inline int f(const int x);

inline int f(const int x)
{
    return 2*x;
}

int main(int argc, char* argv[])
{
    return f(std::atoi(argv[1]));
}

and the constexpr equivalent version :

// Constexpr specifier version
#include<iostream>
#include<cstdlib>

constexpr int f(const int x);

constexpr int f(const int x)
{
    return 2*x;
}

int main(int argc, char* argv[])
{
    return f(std::atoi(argv[1]));
}

My question is : does the constexpr specifier imply the inline specifier in the sense that if a non-constant argument is passed to a constexpr function, the compiler will try to inline the function as if the inline specifier was put in its declaration ?

Does the C++11 standard guarantee that ?

Dillman

'[Will] the compiler try to inline the function' is not what the inline specifier does. (Or maybe I misunderstood your phrasing.) – Theran

The inline specifier no longer has anything to do with inlining – Ephrem

The question foots on the wrong assumption that inline is directly related to inlining. So no, the constexpr specifier doesn't imply the inline specifier in that sense, as that sense doesn't exist. – Advertise

Answer 1

Yes ([dcl.constexpr], §7.1.5/2 in the C++11 standard): "constexpr functions and constexpr constructors are implicitly inline (7.1.2)."

Note, however, that the inline specifier really has very little (if any) effect upon whether a compiler is likely to expand a function inline or not. It does, however, affect the one definition rule, and from that perspective, the compiler is required to follow the same rules for a constexpr function as an inline function.

I should also add that regardless of constexpr implying inline, the rules for constexpr functions in C++11 required them to be simple enough that they were often good candidates for inline expansion (the primary exception being those that are recursive). Since then, however, the rules have gotten progressively looser, so constexpr can be applied to substantially larger, more complex functions.

Paddy

Given that the idea is that constant expressions are evaluated at compile time, I suppose most uses of constexpr functions won't cause any code generation at all... – Prepossessing

@KerrekSB constexpr functions are potentially evaluated at compile time. However the C++14 standard is littered with ones which will very likely be called at runtime. For example: std::array<T,N>::at – Lamdin

@Lamdin yes but only the most-reduced form will remain as opcodes though. e.g: the bound checks, will be evaluated at build time, since their code path is const. But the returned value will be *(data+offset) – Takeshi

I suggest editing this to clarify that it's no longer necessarily the case as of C++17 as per the answer below this one. See open-std.org/jtc1/sc22/wg21/docs/papers/2016/p0386r2.pdf – Fetation

Answer 2

constexpr does not imply inline for non-static variables (C++17 inline variables)

While constexpr does imply inline for functions, it does not have that effect for non-static variables, considering C++17 inline variables.

For example, if you take the minimal example I posted at: How do inline variables work? and remove the inline, leaving just constexpr, then the variable gets multiple addresses, which is the main thing inline variables avoid.

constexpr static variables are however implicitly inline.

Minimal example that constexpr implies inline for functions

As mentioned at: https://mcmap.net/q/17461/-does-constexpr-imply-inline the main effect of inline is not to inline but to allow multiple definitions of a function, standard quote at: How can a C++ header file include implementation?

We can observe that by playing with the following example:

main.cpp

#include <cassert>

#include "notmain.hpp"

int main() {
    assert(shared_func() == notmain_func());
}

notmain.hpp

#ifndef NOTMAIN_HPP
#define NOTMAIN_HPP

inline int shared_func() { return 42; }
int notmain_func();

#endif

notmain.cpp

#include "notmain.hpp"

int notmain_func() {
    return shared_func();
}

Compile and run:

g++ -c -ggdb3  -O0 -Wall -Wextra -std=c++11 -pedantic-errors  -o 'notmain.o' 'notmain.cpp' 
g++ -c -ggdb3  -O0 -Wall -Wextra -std=c++11 -pedantic-errors  -o 'main.o' 'main.cpp' 
g++ -ggdb3  -O0 -Wall -Wextra -std=c++11 -pedantic-errors  -o 'main.out' notmain.o main.o
./main.out

If we remove inline from shared_func, link would fail with:

multiple definition of `shared_func()'

because the header gets included into multiple .cpp files.

But if we replace inline with constexpr, then it works again, because constexpr also implies inline.

GCC implements that by marking the symbols as weak on the ELF object files: How can a C++ header file include implementation?

Tested in GCC 8.3.0.

Swab

BTW, a static class member variable declared constexpr is still inline. cppreference.com: A static member variable (but not a namespace-scope variable) declared constexpr is implicitly an inline variable. – Estellaestelle

@Estellaestelle thanks, I hadn't seen that rule, update answer. – Swab

it's not what open-std.org/JTC1/SC22/WG21/docs/papers/2016/p0386r0.pdf says. it says constexpr implies inline for variables. with no mention of a difference between namespace scope of class scope. – Takeshi

>BTW, a static class member variable declared constexpr is still inline. cppreference.com: A static member variable (but not a namespace-scope variable) declared constexpr is implicitly an inline variable. Just to add, it is only since C+17. – Biforate

@ciro-santilli-trump-ban-is-bad If you have a chance, can you correct your statement "constexpr static variables are however implicitly static."? I assume one of those words "static" was meant to be "inline". – Antoninaantonino

A sid question: is #ifndef - #define a recommended pattern over #pragma once? – Virg

@SohailSi see: stackoverflow.com/questions/1143936/… – Swab

@Takeshi No but it is what open-std.org/jtc1/sc22/wg21/docs/papers/2016/p0386r2.pdf says. The ​constexpr​ specifier implies ​ inline only for static data members, not also for namespacescope variables. – Fetation